Paper 2, Section I, 9B

Cosmology | Part II, 2021

(a) The generalised Boltzmann distribution P(p)P(\mathbf{p}) is given by

P(p)=eβ(Epnpμnp)ZpP(\mathbf{p})=\frac{e^{-\beta\left(E_{\mathbf{p}} n_{\mathbf{p}}-\mu n_{\mathbf{p}}\right)}}{\mathcal{Z}_{\mathbf{p}}}

where β=(kBT)1,μ\beta=\left(k_{B} T\right)^{-1}, \mu is the chemical potential,

Zp=npeβ(Epnpμnp),Ep=m2c4+p2c2 and p=p\mathcal{Z}_{\mathbf{p}}=\sum_{n_{\mathbf{p}}} e^{-\beta\left(E_{\mathbf{p}} n_{\mathbf{p}}-\mu n_{\mathbf{p}}\right)}, \quad E_{\mathbf{p}}=\sqrt{m^{2} c^{4}+p^{2} c^{2}} \quad \text { and } \quad p=|\mathbf{p}|

Find the average particle number N(p)\langle N(\mathbf{p})\rangle with momentum p\mathbf{p}, assuming that all particles have rest mass mm and are either

(i) bosons, or

(ii) fermions .

(b) The photon total number density nγn_{\gamma} is given by

nγ=2ζ(3)π23c3(kBT)3n_{\gamma}=\frac{2 \zeta(3)}{\pi^{2} \hbar^{3} c^{3}}\left(k_{B} T\right)^{3}

where ζ(3)1.2\zeta(3) \approx 1.2. Consider now the fractional ionisation of hydrogen

Xe=nene+nHX_{e}=\frac{n_{e}}{n_{e}+n_{H}}

In our universe ne+nH=np+nHηnγn_{e}+n_{H}=n_{p}+n_{H} \approx \eta n_{\gamma}, where η109\eta \sim 10^{-9} is the baryon-to-photon number density. Find an expression for the ratio

1XeXe2\frac{1-X_{e}}{X_{e}^{2}}

in terms of η,(kBT)\eta,\left(k_{B} T\right), the electron mass mem_{e}, the speed of light cc and the ionisation energy of hydrogen I13.6eVI \approx 13.6 \mathrm{eV}.

One might expect neutral hydrogen to form at a temperature kBTIk_{B} T \sim I, but instead in our universe it happens at the much lower temperature kBT0.3eVk_{B} T \approx 0.3 \mathrm{eV}. Briefly explain why this happens.

[You may use without proof the Saha equation

nHne2=(2π2mekBT)3/2eβI\frac{n_{H}}{n_{e}^{2}}=\left(\frac{2 \pi \hbar^{2}}{m_{e} k_{B} T}\right)^{3 / 2} e^{\beta I}

for chemical equilibrium in the reaction e+p+H+γ.]\left.e^{-}+p^{+} \leftrightarrow H+\gamma .\right]

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